$$ \begin{array}{ll}\frac{2^{2 x}-2^{x}}{2^{x}-1} & ext { (b) } \frac{2^{2 x}-1}{2^{x}+1} \ \frac{16-4^{x}}{2^{x}-4} & ext { (c) } \frac{9^{x}-9}{3^{x}-3}\end{array} $$

Question

$$ \begin{array}{ll}\frac{2^{2 x}-2^{x}}{2^{x}-1} & 	ext { (b) } \frac{2^{2 x}-1}{2^{x}+1} \ \frac{16-4^{x}}{2^{x}-4} & 	ext { (c) } \frac{9^{x}-9}{3^{x}-3}\end{array} $$

UpStudy AI · Accepted Answer

Let's simplify each of the given expressions step by step.

---

### (a) $$\displaystyle \frac{2^{2x} - 2^{x}}{2^{x} - 1}$$

**Step 1:** Factor the numerator.

$$
2^{2x} - 2^{x} = 2^{x} \cdot 2^{x} - 2^{x} = 2^{x}(2^{x} - 1)
$$

**Step 2:** Substitute back into the fraction.

$$
\frac{2^{x}(2^{x} - 1)}{2^{x} - 1}
$$

**Step 3:** Cancel out the common term $$(2^{x} - 1)$$.

$$
= 2^{x} \quad \text{(provided $$2^{x} \neq 1$$, i.e., $$x \neq 0$$)}
$$

---

### (b) $$\displaystyle \frac{2^{2x} - 1}{2^{x} + 1}$$

**Step 1:** Recognize the numerator as a difference of squares.

$$
2^{2x} - 1 = (2^{x})^2 - 1^2 = (2^{x} - 1)(2^{x} + 1)
$$

**Step 2:** Substitute back into the fraction.

$$
\frac{(2^{x} - 1)(2^{x} + 1)}{2^{x} + 1}
$$

**Step 3:** Cancel out the common term $$(2^{x} + 1)$$.

$$
= 2^{x} - 1
$$

---

### (c) $$\displaystyle \frac{9^{x} - 9}{3^{x} - 3}$$

**Step 1:** Rewrite the terms with a common base.

$$
9^{x} = (3^2)^{x} = 3^{2x} \quad \text{and} \quad 9 = 3^2
$$

**Step 2:** Factor the numerator as a difference of squares.

$$
3^{2x} - 3^2 = (3^{x})^2 - 3^2 = (3^{x} - 3)(3^{x} + 3)
$$

**Step 3:** Substitute back into the fraction.

$$
\frac{(3^{x} - 3)(3^{x} + 3)}{3^{x} - 3}
$$

**Step 4:** Cancel out the common term $$(3^{x} - 3)$$.

$$
= 3^{x} + 3 \quad \text{(provided $$3^{x} \neq 3$$, i.e., $$x \neq 1$$)}
$$

---

### Additional Expression: $$\displaystyle \frac{16 - 4^{x}}{2^{x} - 4}$$

**Step 1:** Rewrite the terms with a common base.

$$
4^{x} = (2^2)^{x} = 2^{2x} \quad \text{and} \quad 16 = 2^4
$$

**Step 2:** Factor the numerator.

$$
16 - 4^{x} = 2^4 - 2^{2x} = -(2^{2x} - 2^4) = -\left(2^{x} - 4\right)\left(2^{x} + 4\right)
$$

**Step 3:** Substitute back into the fraction.

$$
\frac{-(2^{x} - 4)(2^{x} + 4)}{2^{x} - 4}
$$

**Step 4:** Cancel out the common term $$(2^{x} - 4)$$.

$$
= -\left(2^{x} + 4\right)
$$

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### Summary of Simplified Expressions

1. **(a)** $$\displaystyle \frac{2^{2x} - 2^{x}}{2^{x} - 1} = 2^{x}$$ \quad (for $$x \neq 0$$)
2. **(b)** $$\displaystyle \frac{2^{2x} - 1}{2^{x} + 1} = 2^{x} - 1$$
3. **(c)** $$\displaystyle \frac{9^{x} - 9}{3^{x} - 3} = 3^{x} + 3$$
4. **Additional:** $$\displaystyle \frac{16 - 4^{x}}{2^{x} - 4} = -\left(2^{x} + 4\right)$$

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These simplifications assume that the denominators are not equal to zero, which imposes the following restrictions on $$x$$:

- For **(a)**: $$2^{x} \neq 1 \Rightarrow x \neq 0$$
- For **(c)**: $$3^{x} \neq 3 \Rightarrow x \neq 1$$
- For the **additional expression**: $$2^{x} \neq 4 \Rightarrow x \neq 2$$

If you have any further questions or need additional assistance, feel free to ask!
Simplified expressions: (a) $$ 2^{x} $$ (for $$ x \neq 0 $$) (b) $$ 2^{x} - 1 $$ (c) $$ 3^{x} + 3 $$ (for $$ x \neq 1 $$) Additional Expression: $$ -\left(2^{x} + 4\right) $$ (for $$ x \neq 2 $$)

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Summary of Simplified Expressions

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